Puzzling problems on gravity adi 1 Tam as Tasn 17. August, 2015. 1 - - PowerPoint PPT Presentation

Teaching Physics Innovatively 2015.

Puzzling problems on gravity

Tam´ as Tasn´ adi 1

• 17. August, 2015.

1BME, Institute of Mathematics

Contents

Introduction Instructive problems Enveloping curve of elliptic orbits Conclusion

Contents

Introduction Instructive problems Enveloping curve of elliptic orbits Conclusion

Overview

Simple laws = ⇒ complicated motions Toolkit:

◮ Newton’s gravitational law ◮ Kepler’s laws ◮ Conservation laws (energy, angular momentum, momentum) ◮ Geometry of conic sections

Most of these problems are discussed in the courses preparing for the IPhO.

Overview

Simple laws = ⇒ complicated motions Toolkit:

◮ Newton’s gravitational law ◮ Kepler’s laws ◮ Conservation laws (energy, angular momentum, momentum) ◮ Geometry of conic sections

Most of these problems are discussed in the courses preparing for the IPhO.

Overview

Simple laws = ⇒ complicated motions Toolkit:

◮ Newton’s gravitational law ◮ Kepler’s laws ◮ Conservation laws (energy, angular momentum, momentum) ◮ Geometry of conic sections

Most of these problems are discussed in the courses preparing for the IPhO.

Contents

Introduction Instructive problems Enveloping curve of elliptic orbits Conclusion

Long pendulum

Problem: Find the period T of a mathematical pendulum whose length L is comparable to the radius of the Earth R. (Assumptions: small angular deviation; small distance from the Earth.) Solution:

◮ |T| ≈ |Fgrav| ≈ mg ◮ T and Fgrav are not vertical ◮ Equation of motion (α, β ≪ 1):

mL¨ α = −mgL(α + β), αL = βR

◮ Result:

T = 2π

• L

g R R + L

R L →∞

− → 2π

• L

g

• α

m L g

Conclusion: Approximate separately the magnitude and the direction of a vector(field).

Long pendulum

Problem: Find the period T of a mathematical pendulum whose length L is comparable to the radius of the Earth R. (Assumptions: small angular deviation; small distance from the Earth.) Solution:

◮ |T| ≈ |Fgrav| ≈ mg ◮ T and Fgrav are not vertical ◮ Equation of motion (α, β ≪ 1):

mL¨ α = −mgL(α + β), αL = βR

◮ Result:

T = 2π

• L

g R R + L

R L →∞

− → 2π

• L

g

• α

m L ~ ~ mg β R Earth T Fgrav ~ ~ mg

Conclusion: Approximate separately the magnitude and the direction of a vector(field).

Long pendulum

Problem: Find the period T of a mathematical pendulum whose length L is comparable to the radius of the Earth R. (Assumptions: small angular deviation; small distance from the Earth.) Solution:

◮ |T| ≈ |Fgrav| ≈ mg ◮ T and Fgrav are not vertical ◮ Equation of motion (α, β ≪ 1):

mL¨ α = −mgL(α + β), αL = βR

◮ Result:

T = 2π

• L

g R R + L

R L →∞

− → 2π

• L

g

• α

m L ~ ~ mg β R Earth T Fgrav ~ ~ mg

Conclusion: Approximate separately the magnitude and the direction of a vector(field).

Total energy of elliptic orbits

Problem: An object of mass m is orbiting another object of mass M ≫ m. Express the total me- chanical energy E(a, b) in terms

• f the major and minor axes a and

b.

m a c b a M A P

Solution:

◮ Distance of perihelion P and aphelion A:

rP = a − c, rA = a + c, a2 = b2 + c2

◮ Energy conservation: E(a, b) = mv2

P

2

− G mM

rP = mv2

A

2

− G mM

rA ◮ Angular momentum conservation: mvPrP = mvArA ◮ Result:

E(a) = −mMG 2a Conclusion: The total energy of an elliptic orbit depends only on the major axis a.

Total energy of elliptic orbits

Problem: An object of mass m is orbiting another object of mass M ≫ m. Express the total me- chanical energy E(a, b) in terms

• f the major and minor axes a and

b.

m a c b a P A M

A

r rP

Solution:

◮ Distance of perihelion P and aphelion A:

rP = a − c, rA = a + c, a2 = b2 + c2

◮ Energy conservation: E(a, b) = mv2

P

2

− G mM

rP = mv2

A

2

− G mM

rA ◮ Angular momentum conservation: mvPrP = mvArA ◮ Result:

E(a) = −mMG 2a Conclusion: The total energy of an elliptic orbit depends only on the major axis a.

Total energy of elliptic orbits

Problem: An object of mass m is orbiting another object of mass M ≫ m. Express the total me- chanical energy E(a, b) in terms

• f the major and minor axes a and

b.

m a c b a P A M

A

r rP

Solution:

◮ Distance of perihelion P and aphelion A:

rP = a − c, rA = a + c, a2 = b2 + c2

◮ Energy conservation: E(a, b) = mv2

P

2

− G mM

rP = mv2

A

2

− G mM

rA ◮ Angular momentum conservation: mvPrP = mvArA ◮ Result:

E(a) = −mMG 2a Conclusion: The total energy of an elliptic orbit depends only on the major axis a.

Deviation angle of hyperbolic orbits

Problem: A comet passes by the Sun. Determine its angle of deviation α in terms of the initial speed v0 and the im- pact parameter p.

v0 α p

Hint:

◮ Apply energy and angular momentum conservation for the

perihelion P and the point at infinity

◮ Use the geometry of the hyperbola:

c2 = a2 + b2, tan α 2

• = b

a, p = a, PF = c − b

Deviation angle of hyperbolic orbits

Problem: A comet passes by the Sun. Determine its angle of deviation α in terms of the initial speed v0 and the im- pact parameter p.

c a b α α/2 P F

Hint:

◮ Apply energy and angular momentum conservation for the

perihelion P and the point at infinity

◮ Use the geometry of the hyperbola:

c2 = a2 + b2, tan α 2

• = b

a, p = a, PF = c − b

Racing satellites

Problem: Two satellites, A and B or- bit the Earth on the same circular orbit, B lags behind A. How should B use its rocket in order to catch up with A? (Assume that the rocket can give only a quick impulse to the satellite.)

B A Earth

Solution:

◮ If B increases its speed =

⇒ E(a) = − mMG

2a

< 0 increases = ⇒ a increases = ⇒ period T increases WRONG!!!

◮ If B decreases its speed =

⇒ E(a) decreases = ⇒ a decreases = ⇒ T decreases CORRECT!!!

Racing satellites

Problem: Two satellites, A and B or- bit the Earth on the same circular orbit, B lags behind A. How should B use its rocket in order to catch up with A? (Assume that the rocket can give only a quick impulse to the satellite.)

B A E F

Solution:

◮ If B increases its speed =

⇒ E(a) = − mMG

2a

< 0 increases = ⇒ a increases = ⇒ period T increases WRONG!!!

◮ If B decreases its speed =

⇒ E(a) decreases = ⇒ a decreases = ⇒ T decreases CORRECT!!!

Racing satellites

Problem: Two satellites, A and B or- bit the Earth on the same circular orbit, B lags behind A. How should B use its rocket in order to catch up with A? (Assume that the rocket can give only a quick impulse to the satellite.)

B A E F

Solution:

◮ If B increases its speed =

⇒ E(a) = − mMG

2a

< 0 increases = ⇒ a increases = ⇒ period T increases WRONG!!!

◮ If B decreases its speed =

⇒ E(a) decreases = ⇒ a decreases = ⇒ T decreases CORRECT!!!

Stopping the Moon

Problem: Imagine that the Moon’s orbital motion around the Earth is suddenly stopped. How long would it take for the Moon to fall into the Earth? Remark: The direct integration is beyond the secondary school level. Idea: Apply Kepler’s third law for the two orbits of the Moon.

Stopping the Moon

Problem: Imagine that the Moon’s orbital motion around the Earth is suddenly stopped. How long would it take for the Moon to fall into the Earth? Remark: The direct integration is beyond the secondary school level. Idea: Apply Kepler’s third law for the two orbits of the Moon.

Stopping the Moon

Problem: Imagine that the Moon’s orbital motion around the Earth is suddenly stopped. How long would it take for the Moon to fall into the Earth? Remark: The direct integration is beyond the secondary school level. Idea: Apply Kepler’s third law for the two orbits of the Moon.

Motions observed from a space station

Problem: A space station is orbit- ing the Earth on a circular trajec- tory, facing always with the same side towards the Earth. A small

• bject is thrown out of the space

station with a small initial veloc- ity. How does the object move relative to the space station?

R x y ω y x Earth space station

Remarks:

◮ Solve the problem in rotating reference frame ◮ Approximate the magnitude and the direction of the force

vectors separately

Motions observed from a space station

Problem: A space station is orbit- ing the Earth on a circular trajec- tory, facing always with the same side towards the Earth. A small

• bject is thrown out of the space

station with a small initial veloc- ity. How does the object move relative to the space station?

R x y ω y x Earth space station

Remarks:

◮ Solve the problem in rotating reference frame ◮ Approximate the magnitude and the direction of the force

vectors separately

Contents

Introduction Instructive problems Enveloping curve of elliptic orbits Conclusion

The problem

Let A be a fixed point in space at a distance d from a fixed sun S

• f mass M. Particles of mass m are shot from A in different

directions at constant speed v. Which points can be reached by the particles?

S M A v m d

The enveloping curve of the orbits is to be found.

First notices

◮ Planar trajectories, rotational symmetry about AS =

⇒ restrict attention to the plane

◮ Total energy: E = 1 2mv2 − G mM d

is constant

◮ If E ≥ 0 =

⇒ infinite orbits = ⇒ any point can be reached (proof is omitted now)

◮ If E < 0 =

⇒ finite orbits = ⇒ only a bounded region can be reached (this case is studied now)

◮ The semi-major axis a of the orbits:

E = 1 2mv2 − G mM d = −mMG 2a = ⇒ a = MG 2MG − dv2 d

◮ 2a > d

Enveloping curve

Question: How to obtain the enveloping curve of a family of curves {Cα}α∈I? Answer:

◮ Two curves close to each

• ther: Cα and Cβ

◮ The intersection K of Cα

and Cβ is close to the enveloping curve

◮ The point Pα where Cα

touches the enveloping curve is: Pα = lim

β→α Cα ∩ Cβ

• K

curve enveloping

α

C Cβ

Enveloping curve

Question: How to obtain the enveloping curve of a family of curves {Cα}α∈I? Answer:

◮ Two curves close to each

• ther: Cα and Cβ

◮ The intersection K of Cα

and Cβ is close to the enveloping curve

◮ The point Pα where Cα

touches the enveloping curve is: Pα = lim

β→α Cα ∩ Cβ

• K

curve enveloping Cα C Pα K

β

Solution (Enveloping curve of elliptic orbits)

◮ Focal points of an orbit: S and Fα

AS + AFα = 2a = ⇒ AFα = 2a − d (circle)

◮ SK + KFα = SK + KFβ = 2a =

⇒ KFα = KFβ = ⇒ A, Fα, Pα are collinear

◮ SPα + PαA = SPα + PαFα

• 2a

+ FαA

• 2a−d

= 4a − d; ellipse

Fα S A a−d 2 d

Solution (Enveloping curve of elliptic orbits)

◮ Focal points of an orbit: S and Fα

AS + AFα = 2a = ⇒ AFα = 2a − d (circle)

◮ SK + KFα = SK + KFβ = 2a =

⇒ KFα = KFβ = ⇒ A, Fα, Pα are collinear

◮ SPα + PαA = SPα + PαFα

• 2a

+ FαA

• 2a−d

= 4a − d; ellipse

Fα Pα Pβ Fβ S A K

Solution (Enveloping curve of elliptic orbits)

◮ Focal points of an orbit: S and Fα

AS + AFα = 2a = ⇒ AFα = 2a − d (circle)

◮ SK + KFα = SK + KFβ = 2a =

⇒ KFα = KFβ = ⇒ A, Fα, Pα are collinear

◮ SPα + PαA = SPα + PαFα

• 2a

+ FαA

• 2a−d

= 4a − d; ellipse

Pα Fα S A

Solution (Enveloping curve of elliptic orbits)

◮ Focal points of an orbit: S and Fα

AS + AFα = 2a = ⇒ AFα = 2a − d (circle)

◮ SK + KFα = SK + KFβ = 2a =

⇒ KFα = KFβ = ⇒ A, Fα, Pα are collinear

◮ SPα + PαA = SPα + PαFα

• 2a

+ FαA

• 2a−d

= 4a − d; ellipse

Pα Fα S A

Enveloping curve is an ellipse of foci S and A, semi-major axis 2a− d

2 .

The orbits

• 2.5
• 2
• 1.5
• 1
• 0.5

0.5 1 1.5 2 2.5

• 2
• 1.5
• 1
• 0.5

0.5 1 1.5 2 "fifi.dat"

Contents

Introduction Instructive problems Enveloping curve of elliptic orbits Conclusion

What can students learn from puzzling problems on gravity?

◮ The laws of gravity ◮ Conservation laws (energy, momentum, angular momentum) ◮ Geometry of conic sections ◮ Approximation techniques ◮ Small perturbations ◮ Changing reference frames ◮ Inertial forces ◮ Enveloping curves ◮ . . .

Thank you for your attention!

What can students learn from puzzling problems on gravity?

◮ The laws of gravity ◮ Conservation laws (energy, momentum, angular momentum) ◮ Geometry of conic sections ◮ Approximation techniques ◮ Small perturbations ◮ Changing reference frames ◮ Inertial forces ◮ Enveloping curves ◮ . . .

Thank you for your attention!